# Mean of a practical distribution

I have a graph with an asymmetrical distribution (spectral response for some sensor). The graph is plotted as efficiency values versus vavelength. I must determine the median wavelength. Help please, my statistics is so rusty !
I have determined the mean value of the efficiency - the values on the y-axis, and considered selecting the wavelength (x-value) corresponding to that value - but that doesn't seem to give me a relevant answer, and it is not even close to the center of the plot (even in the case where it should be). I thought of cheating and just getting the median of the graph - but the graphs are weird. In one case, the graph has a fairly rectangular shape (rise, sort of plateau but with mountains and valleys, then fall), in another case, there is a skewed peak on the left and another tiny bump on the right.
I would appreciate any suggestions ! Thank you !

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Do you want the median or the mean? How did you determine the mean of the efficency? Do you have a sample? – Xabier Domínguez May 1 '12 at 16:43
Used Excel's mean, and noticed that it was wrong. – Thalia May 2 '12 at 17:57

I was eventually able to find the mean as $\dfrac{\sum (p_i x_i)} { \sum (p_ i)}$ .
I am not even sure -- What $\pi$ and $x_i$ are? Why don't you tell us what you wanted to find? What you are given? It is not even clear, even from the question. – user21436 May 4 '12 at 12:35
I have now edited it accordingly. But, if you wanted $p_i$ then, you would atleast write p_i and not pi. Looking around the site will tell you this site allows TeX mark up. Go through my edit by clicking on the timestamp available above my name. I have now removed my downvote. (FWIW, I do know what the expected value of a random variable is, but still feel that you should write your answer more clearly.) Regards, – user21436 May 5 '12 at 16:09